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Suppose I want to solve $\int_{-1}^1 u(x)^{\ 2}\ dx$. I should be able to write this as $\int_{-1}^1 u^{\ 2} \frac{1}{u'}\ du$.

Since $\int_{-1}^1 u'\ du$ is presumably just $u(1) - u(-1)$, can I also find the definite integral of an integrand in $u$ and $u'$, such as $\int_{-1}^1 u^{\ 2} \frac{1}{u'}\ du$? How would I proceed in doing so?

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  • $\begingroup$ you can substitute like that only if $u'$ is constant. otherwise you would still have a function of x in the integral. you also have to interchange limits of integration, so the integral would be from $u(-1)$ to $u(1)$. $\endgroup$ – Yizhar Amir Mar 26 at 16:43
  • $\begingroup$ Unless $u(1)=1$ and $u(-1)=-1$ and $u(x)$ is increasing in the interval, your substitution won't be valid. $\endgroup$ – herb steinberg Mar 26 at 17:32
  • $\begingroup$ @YizharAmir Isn't what I did very similar to youtu.be/_60sKaoRmhU?t=840 I just multiplied the $dx$ by $\frac{du}{du}$ to get $\frac{dx}{du}\ du \implies \frac{1}{u'}\ du$. $\endgroup$ – user10478 Mar 27 at 3:48

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